(1) Let \(H\)  be a compact subgroup of \(G\) . We then have \(\mathcal {C}_c(G)=\mathcal {C}_c(G|H)\) . To see this, let \(\psi \in \mathcal {C}_c(G|H)\)  be given. Then \(\mathrm {q}({\rm supp}(\psi ))\)  is compact in \(G/H\) . Using Lemma 2.46 of [1]}, there exists a compact subset \(F\)  of \(G\)  such that \(\mathrm {q}(F)=\mathrm {q}({\rm supp}(\psi ))\) . This implies that \(\mathrm {supp}(\psi )\subseteq FH\) . Since \(H\)  is compact, we conclude that \(\mathrm {supp}(\psi )\)  is compact in \(G\)  and so \(\psi \in \mathcal {C}_c(G)\) . Therefore, we deduce that \(M_\xi (G,H)\subseteq \mathcal {C}_c(G)\) .
(2) Let \(f\in \mathcal {C}_c(G)\)  be given. Using (1), we have \(T_\xi (f)\in \mathcal {C}_c(G)\) . Then, for each \(x\in G\)  and since \(\lambda _H\)  is a probability measure, we get
\(T_\xi (T_\xi (f))(x)&=\int _HT_\xi (f)(xs)\overline{\xi (s)}\mathrm {d}\lambda _H(s)\\&=\int _HT_\xi (f)(x)\xi (s)\overline{\xi (s)}\mathrm {d}\lambda _H(s)\\&=T_\xi (f)(x)\int _H|\xi (s)|^2\mathrm {d}\lambda _H(s)=T_\xi (f)(x)\left(\int _H\mathrm {d}\lambda _H(s)\right)=T_\xi (f)(x).\) 
